I'm working through Browder's Mathematical Analysis (http://www.springer.com/in/book/9780387946146) and am having a bit of trouble being convinced by one of his definitions.
On Page 288, for Definition 13.5, he says: Let $\omega_0 = dx^1 \wedge \dots \wedge dx^n$; for each $j, 1\leq j \leq n$, we put $$\eta^j = (-1)^j dx^1 \wedge dx^{j-1} \wedge dx^{j+1} \dots \wedge dx^n$$
The $(-1)^j$ is justified so that $dx^j \wedge \eta^j = \omega_0$.
My doubt is that: Shouldn't it be $(-1)^{j-1}$?
I test Browder's definition with the $j =1, 2$ cases to get: $$dx^1 \wedge (-1)^1 dx^2\wedge \dots \wedge dx^n = -\omega_0 $$ and $$dx^2 \wedge (-1)^2 dx^1\wedge dx^3 \wedge \dots \wedge dx^n = -\omega_0$$
Each time, I swap positions of $dx^i$s, I multiply by $(-1)$. In the first case, no swapping was required and in the second case, one swap was required. Essentially, what I'm asserting is that in the general case, $j-1$ swaps are required so why are we multiplying the $\eta^j$ expression by $(-1)^j$ instead of $(-1)^{j-1}$?